What is the radius and center of the image of $|z|=1$ under $f(z) = \frac{3z+2}{4z+3}$?

I would like to compute the image of the circle $$|z|=1$$ about the fractional linear transformation: $$f(z) = \frac{3z+2}{4z+3}$$ In particular, I'd like to compute the new center and radius.

The Möbius transformation can be turned into inversion as well:

• $$C_1= 4|z|^2+3\overline{z}-3z-2$$
• $$C_2 =|z|^2 - 1$$

Or we could turn the second circle into a fractional lineartrasnforation $$g(z) = - \frac{1}{z}$$. Then I could multiply the two transformations: $$\left[ \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right] \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 2 & 3 \\ 3 & 4 \end{array} \right]$$

and this could turn back into a circle:

• $$C_1C_2 = 3|z|^2 + 4 \overline{z} + 3z + 2$$

I found this technique in an somewhat dated geometry textbook from the 1930's and I'm still figuring out their notation. I definitely like the idea that Möbius transformations and Circles can be identified.

• Why is accepted solution more usefull than mine? – Aqua Jan 12 at 9:58

I would take three points on $$\mid z\mid=1$$ and see where they go. As noted in @greedoid's answer, we have $$f(1)=\frac57\,,f(-1)=1$$ and $$f(i)=\frac{18+i}{25}$$.

Since these points are not colinear, the image is indeed a circle.

So, if $$z$$ is the center, we have: $$\mid z-1\mid=\mid z-\frac57\mid=\mid z-\frac{18+i}{25}\mid=r$$.

This leads via a little algebra to $$z=\frac67$$. Thus $$r=\frac17$$.

Here is an automatic procedure: first invert the relation $$w=f(z)$$, then apply the condition $$|z|=1$$ to the inverse formula $$z=g(w)$$ to deduce an equation of the image set.

In the present case, $$w=f(z)$$ means that $$w=\frac{3z+2}{4z+3}$$ that is, $$(4z+3)w=3z+2$$, that is, $$(4w-3)z=2-3w$$, that is, $$z=\frac{2-3w}{4w-3}$$ Thus, the image of the circle has equation $$\left|\frac{2-3w}{4w-3}\right|=1$$ In turn, this means successively that $$|2-3w|=|4w-3|$$ that is, $$|2-3w|^2=|4w-3|^2$$ that is, $$4-6(w+\bar w)+9|w|^2=16|w|^2-12(w+\bar w)+9$$ that is, $$7|w|^2-6(w+\bar w)+5=0$$ and finally, if $$w=x+iy$$, $$7(x^2+y^2)-12x+5=0$$ from which you might be able to conclude that the desired radius is $$r=\frac17$$ As one can see, switching to the decomposition of complex numbers into their real and imaginary parts as late as possible in the computations, simplifies these.

Edit: The comment by user @alex.jordan below shows eloquently that "as late as possible" just above, could even be replaced by "never"...

• +1 You can avoid $x$ and $y$ entirely. Dividing the line right before $x$ and $y$ come in by $7$ gives $|w|^2-\frac67(w+\bar{w})+\frac57=0$. Now "complete the square" to get $|w|^2-\frac67(w+\bar{w})+\frac{36}{49}=\frac{36}{49}-\frac57=\frac1{49}$. That's $\left(w-\frac67\right)\left(\bar{w}-\frac67\right)=\frac1{49}$. Now take the square root: $|w-\frac67|=\frac17$. – alex.jordan Jan 11 at 6:34
• @alex.jordan Indeed. Well done and thanks. – Did Jan 11 at 9:02

Let see where this transformation takes $$1,-1$$ and $$i$$:

$$\begin{eqnarray} 1&\longmapsto &{5\over 7}\\-1&\longmapsto &1\\i&\longmapsto &{18+i\over 25}\\ \end{eqnarray}$$

Now calculate the center and radius of a triangle on $$\alpha ={5\over 7}$$, $$\beta =1$$ and $$\gamma ={18+i\over 25}$$.

Since this triangle is right at $$\gamma$$ we see that midpoint of segment $$\alpha \beta$$, that is $$\sigma = {6\over 7}$$ is a center of new circle with $$r = {1\over 7}$$.

Using Inversive Geometry

For a given LFT $$\frac{az+b}{cz+d}$$ and circle of radius $$r$$ centered at $$k$$, the antipodal points of that source circle $$k\pm\frac{k+d/c}{|k+d/c|}r\tag1$$ get mapped by the LFT to antipodal points of the image circle.

This is because these points are on the line containing the center of the circle, $$k$$, and the center of the inversion, $$-d/c$$. Any line through the center of the inversion is mapped to a line, and since that line is perpendicular to the source circle at the points of intersection, the image line is perpendicular to the image circle; that is, they intersect at antipodal points.

If $$c=0$$ (the LFT is simply affine) or $$k+d/c=0$$ (the center of the source circle is the center of the inversion), then any two antipodal points get mapped to antipodal points, so replace $$\frac{k+d/c}{|k+d/c|}$$ with any point on the unit circle in $$\mathbb{C}$$.

If one of the points computed in $$(1)$$ equals $$-\frac dc$$ (that is, that point is mapped to $$\infty$$ by the LFT), then the circle is mapped to a line. In that case, just plug any other two points on the source circle into the LFT to get two points on the image line.

Given a pair of antipodal points on a circle, $$\{p_1,p_2\}$$, the radius, r, and center, k, of that circle are given by $$r=\frac{|p_1-p_2|}2\qquad k=\frac{p_1+p_2}2\tag2$$

Application

In this case, we have $$\frac{3z+2}{4z+3}$$, $$k=0$$, and $$r=1$$. Therefore, $$(1)$$ says that the antipodal points of the source circle $$0\pm\frac{0+3/4}{|0+3/4|}\cdot1=\{-1,1\}\tag3$$ get mapped by the LFT to the antipodal points of the image circle $$\left\{1,\frac57\right\}\tag4$$ then $$(2)$$ says that the radius, $$r$$, and the center, $$k$$, of the image circle are $$\bbox[5px,border:2px solid #C0A000]{r=\frac17\qquad k=\frac67}\tag5$$

The image circle is symmetric wrt the real axis, therefore $$[f(1), f(-1)] = [5/7, 1]$$ is a diameter.

Here's another solution I was able to find. Notice the matrix factorization:

$$\left[ \begin{array}{cc} 3 & 2 \\ 4 & 3 \end{array} \right] = \left[ \begin{array}{cc} \frac{1}{5} & 0\\ 0 & 5 \end{array} \right] \times \left[ \begin{array}{cc} 1 & 18 \\ 0 & 1 \end{array} \right] \times \left[ \begin{array}{cr} \frac{3}{5}& -\frac{4}{5}\\ \frac{4}{5}& \frac{3}{5} \end{array} \right] = A \times B \times C$$ The geometry behind this is that we have a Möbius transformation that factors into three parts: $$\text{Möbius} = rotation \times translation \times dilation$$

Now we have that $$|z|=1$$ is a circle centered at the origin passing through the points $$z = \pm 1$$ and $$z = i$$. In fact, all these transformation will map to circles symmetric about the real axis. Here are the endpoints after the respective transformations:

$$(-1,1) \stackrel{C}{\to} (7, - \frac{1}{7}) \stackrel{B}{\to} (25,\frac{125}{7})\stackrel{A}{\to} (1, \frac{5}{7})$$ This corresponds to a circle centered at $$z = \frac{6}{7}$$ with radius $$\frac{1}{7}$$.

One possibility for computing this image circle is to notice the circle $$|z|=1$$ is a geodesic curve in the upper-half plane (with metric $$ds^2 = \frac{dx^2 +dy^2}{y^2}$$) and passing through the point $$(z, \vec{u}) = (i, (1,0)) \in T_1(\mathbb{H})$$.

A Möbius transformation on $$\mathbb{H}$$ can be "lifted" to a Möbius transformation on $$T_1(\mathbb{H})$$ like this: $$\left[ z \mapsto \frac{az+b}{cz+d} \right] \to \left[ (z, \vec{u}) \mapsto \left( \frac{az+b}{cz+d} , \frac{\vec{u}}{(cz+d)^2} \right) \right]$$ Let's see what happens when I try the previous example here: $$\big(i, (1,0)\big) \mapsto \left( \frac{3i+2}{4i+3}, \frac{(1,0)}{(4i+3)^2}\right) = \left( \frac{18+i}{25} , \frac{1}{25}(24,-7) \right)$$ The factor of $$\frac{1}{25}$$ can be discarded since we only need the unit vector. This map is an isometry in hyperbolic space. The vector $$\vec{u}$$ would be tangent to a semi-circle with radius in the direction $$\vec{u}_\perp$$ passing through the point $$f(z)=(\frac{18}{25}, \frac{1}{25})$$. Therefore the center would be: $$(\frac{18}{25}, \frac{1}{25}) + \frac{1}{7 \times 25}(24,-7) = (\frac{1}{7},0)$$ agreeing with the previous answer.